Differentiation (AI HL)
Differentiation finds the rate of change of a function - its gradient at any point. At AI HL this means differentiating a wider set of functions than at SL, using the chain, product and quotient rules, then applying that gradient function to real problems: finding stationary points, testing whether they're a maximum or minimum, optimising a quantity like cost or volume, and tracking related rates of change. It's one of the most heavily examined topics on both papers.
What the syllabus says
This topic draws on the SL calculus content (examinable at HL) plus the AHL extension points on differentiation.
| Code | Syllabus content |
|---|---|
| SL5.3 | The derivative of \(f(x)=ax^n\) is \(f'(x)=anx^{n-1}\), for functions of the form \(f(x)=ax^n+bx^{n-1}+\dots\) where all exponents are integers. |
| SL5.4 | Tangents and normals at a given point, and their equations, using both analytic approaches and technology. |
| SL5.6 | Values of \(x\) where the gradient of a curve is zero. Solution of \(f'(x)=0\). Local maximum and minimum points, and awareness that a local extremum need not be the greatest or least value on the whole domain. |
| SL5.7 | Optimisation problems in context, such as maximising profit, minimising cost, or maximising volume for a given surface area. |
| AHL5.9 | The derivatives of \(\sin x\), \(\cos x\), \(\tan x\), \(e^x\), \(\ln x\), \(x^n\) where \(n\in\mathbb{R}\). The chain rule, product rule and quotient rule. Related rates of change. |
| AHL5.10 | The second derivative, using both \(\dfrac{d^2y}{dx^2}\) and \(f''(x)\) notation. Use of the second derivative test to distinguish a maximum from a minimum. The terms "concave-up" (\(f''(x)>0\)) and "concave-down" (\(f''(x)<0\)), and interpretation of a point of inflexion as a point where concavity changes. |
SL5.3, SL5.4, SL5.6 and SL5.7 are core content shared with AA, examinable at AI HL alongside the AHL5.9-5.10 extension.
Key terms
Five words worth knowing cold before you touch the formulas below - each with a worked example showing exactly what it means.
What is a derivative?
The derivative of a function tells you its gradient at any point - how fast \(y\) is changing as \(x\) changes. It's found using differentiation rules and written \(f'(x)\) or \(\dfrac{dy}{dx}\). A steep positive derivative means the function is rising quickly; a negative one means it's falling.
e.g. For \(f(x)=x^3\), \(f'(x)=3x^2\), so \(f'(2)=3(2)^2=12\).
What is a stationary point?
A stationary point is any point on a curve where the gradient is exactly zero - the tangent is horizontal. You find one by solving \(f'(x)=0\). It could be a local maximum, a local minimum, or (less commonly) a point of inflexion.
e.g. For \(f(x)=x^2-6x+5\), \(f'(x)=2x-6=0 \Rightarrow x=3\), giving the point \((3,-4)\).
What is the second derivative test?
The second derivative test classifies a stationary point by checking the sign of \(f''(x)\) there. A negative value means the curve is concave-down (a maximum); a positive value means concave-up (a minimum). It's usually faster than testing gradient signs either side.
e.g. For \(f(x)=x^3-6x^2+9x\), \(f''(x)=6x-12\); at \(x=1\), \(f''(1)=-6<0\), so \((1,4)\) is a maximum.
What is the chain rule?
The chain rule differentiates a "function of a function" - anything where one expression is nested inside another. Differentiate the outer function first, leaving the inner one unchanged, then multiply by the derivative of the inner function.
e.g. For \(y=(3x+1)^4\), \(\dfrac{dy}{dx}=4(3x+1)^3\times3=12(3x+1)^3\), so at \(x=0\), \(\dfrac{dy}{dx}=12\).
What is optimisation?
Optimisation uses differentiation to find the maximum or minimum value of a real quantity - area, cost, volume - subject to some constraint. You write the quantity as a function of a single variable, differentiate, set the derivative to zero, and check the nature of the resulting stationary point.
e.g. A rectangle has perimeter 20: \(A=x(10-x)=10x-x^2\), \(\dfrac{dA}{dx}=10-2x=0 \Rightarrow x=5\), giving maximum area \(25\).
Key formulas
A handful of differentiation rules cover almost every question on this topic. The tables below summarise them at a glance - the explanations underneath go into more depth on each one.
Formula reference
The standard derivatives and the chain, product and quotient rules are on the formula booklet; the method for classifying a stationary point is a technique, not a listed formula.
| Formula | Used for | Booklet? |
|---|---|---|
| \(\dfrac{d}{dx}(x^n)=nx^{n-1}\) | Power rule, including negative/fractional \(n\) | ✓ Yes |
| \(\dfrac{d}{dx}(\sin x)=\cos x,\ \dfrac{d}{dx}(\cos x)=-\sin x,\ \dfrac{d}{dx}(e^x)=e^x,\ \dfrac{d}{dx}(\ln x)=\dfrac1x\) | Standard derivatives | ✓ Yes |
| \(\dfrac{dy}{dx}=\dfrac{dy}{du}\cdot\dfrac{du}{dx}\) | Chain rule | ✓ Yes |
| \((uv)' = u'v+uv'\) | Product rule | ✓ Yes |
| \(\left(\dfrac{u}{v}\right)' = \dfrac{u'v-uv'}{v^2}\) | Quotient rule | ✓ Yes |
| Solve \(f'(x)=0\), then check the sign of \(f''(x)\) | Finding and classifying stationary points | Not in booklet - technique, not a formula |
First derivative vs second derivative
The first and second derivatives tell you different things about the same curve - one about slope, one about curvature.
| Feature | First derivative \(f'(x)\) | Second derivative \(f''(x)\) |
|---|---|---|
| Measures | Gradient of the curve | Rate of change of the gradient |
| Zero means | A stationary point | A possible point of inflexion |
| Positive means | Curve is increasing | Curve is concave-up |
| Negative means | Curve is decreasing | Curve is concave-down |
| Notation | \(\dfrac{dy}{dx}\) or \(f'(x)\) | \(\dfrac{d^2y}{dx^2}\) or \(f''(x)\) |
Rules for differentiating
Which rule you need depends on the structure of the expression, not just which functions appear in it.
Power rule
\[\dfrac{d}{dx}(x^n)=nx^{n-1}\]
Multiply by the exponent, then reduce the exponent by 1. Works for negative and fractional \(n\) too.
✓ In the formula bookletChain rule
\[\dfrac{dy}{dx}=\dfrac{dy}{du}\cdot\dfrac{du}{dx}\]
For a function nested inside another - differentiate the outside, keep the inside, then multiply by the derivative of the inside.
✓ In the formula bookletProduct rule
\[(uv)' = u'v+uv'\]
For two functions multiplied together - differentiate each in turn, keeping the other unchanged, and add the results.
✓ In the formula bookletQuotient rule
\[\left(\dfrac{u}{v}\right)' = \dfrac{u'v-uv'}{v^2}\]
For one function divided by another - the order in the numerator matters, and always divide by \(v^2\).
✓ In the formula bookletStationary points, second derivative and optimisation
These three ideas build on each other: find where \(f'(x)=0\), classify the point, then apply the same method to a real-world quantity.
Finding stationary points
Differentiate, set \(f'(x)=0\), and solve for \(x\).
Substitute each \(x\)-value back into \(f(x)\) to get the full coordinate pair.
Not in the formula booklet - method, not a formulaClassifying with \(f''(x)\)
\(f''(x)<0 \Rightarrow\) maximum. \(f''(x)>0 \Rightarrow\) minimum.
If \(f''(x)=0\), the test is inconclusive - check the sign of \(f'(x)\) either side instead.
Not in the formula booklet - techniqueOptimisation method
Write the quantity as a function of one variable, using a constraint to eliminate the other.
Differentiate, solve \(f'(x)=0\), confirm it's the required max/min, then answer the question actually asked.
Not in the formula booklet - applied methodWorked examples
Two full exam-style questions, marked exactly like the real thing. Try each one yourself before checking the worked solution.
For \(y = x^3 - 6x^2 + 9x.\)
(a)(i) Find the coordinates of the stationary point with the smaller \(x\)-value.
(a)(ii) Find the coordinates of the stationary point with the larger \(x\)-value.
(b)(i) Determine the nature of the stationary point with the smaller \(x\)-value, using the second derivative.
(b)(ii) Determine the nature of the stationary point with the larger \(x\)-value.
Worked solution
(a)(i) \(\dfrac{dy}{dx} = 3x^2 - 12x + 9 = 0.\) M1
\(3x^2 - 12x + 9 = 3(x-1)(x-3) = 0.\) A1
Point \((1, 4).\) A1
(a)(ii) Point \((3, 0).\) A1
(b)(i) \(\dfrac{d^2y}{dx^2} = 6x - 12.\) M1
At \(x=1\): \(-6 < 0\) → max. A1
(b)(ii) At \(x=3\): \(6 > 0\) → min. A1
A farmer has 200 m of fencing to make a rectangular enclosure against a straight wall (no fence needed along the wall). Let the two sides perpendicular to the wall have length \(x\).
(a) Show that the area is \(A = 200x - 2x^2.\)
(b) Find the value of \(x\) that maximises the area.
(c) Find the maximum area.
Worked solution
(a) Side parallel to wall \(= 200 - 2x.\) Area \(A = x(200 - 2x)\) M1
\(= 200x - 2x^2.\) A1 AG
(b) \(\dfrac{dA}{dx} = 200 - 4x = 0 \Rightarrow x\) M1 \(= 50\) m. A1 (\(A''=-4<0\), max.) A1
(c) \(A = 200(50) - 2(50)^2\) M1
\(= 5000\) m². A1
Common mistakes
The four slip-ups that account for most of the marks lost on this topic - worth reading before you start practising, not just after you get one wrong.
- Forgetting the chain rule on composite functions. Differentiating \(\sin(3x)\) as \(\cos(3x)\) instead of \(3\cos(3x)\) drops the "multiply by the derivative of the inside" step - always check whether the argument of the function is anything other than plain \(x\).
- Reporting the gradient instead of the coordinate. A stationary point is a pair \((x,y)\), not just the \(x\)-value that solves \(f'(x)=0\) - you still need to substitute back into \(f(x)\) to find \(y\).
- Assuming every solution of \(f'(x)=0\) is a maximum or minimum. Some are points of inflexion where \(f''(x)=0\) too, and the second derivative test needs a different check (the sign of \(f'(x)\) either side) whenever \(f''(x)=0\).
- Sign errors substituting negative values into \(f''(x)\). A careless \((-2)^2\) becoming \(-4\) instead of \(4\) flips a minimum into a "maximum" - substitute carefully and use brackets.
Using your GDC
Every step below is a real button sequence, not a vague "use your calculator" hint - covering the TI-84 Plus, TI-Nspire, and Casio fx-9860/fx-CG50. Pick your model to filter down to just the steps that apply to you.
Gives a gradient instantly to check your differentiation or when a function is awkward.
- MATH → 8:nDeriv(, then enter nDeriv(f(x), x, a). Or graph and use 2nd → CALC → 6:dy/dx.TI-84
- menu → Calculus → Numerical Derivative at a Point.Nspire
- Run-Matrix → MATH (F4) → d/dx, then enter the function and the x-value.Casio
- Read off the gradient - useful for tangent slopes without algebra.
Tip: Handy for checking the gradient at a point or finding a tangent's slope.
See the full GDC guide for more calculator models and topics.
Ready to practise properly?
Differentiation questions, marked instantly like the real exam.
Quick answers
The questions students on this topic ask most often.
How do I know if a stationary point is a maximum or a minimum?
Use the second derivative test: find \(f''(x)\) and substitute the \(x\)-value of the stationary point. If \(f''(x) < 0\) the point is a local maximum; if \(f''(x) > 0\) it's a local minimum. If \(f''(x) = 0\) the test is inconclusive and you need another method.
What's the difference between differentiation and optimisation?
Differentiation is the technique - finding a derivative. Optimisation is an application of it: you write a quantity like area, cost or volume as a function of one variable, differentiate, set the derivative to zero to find the stationary point, then check whether it's a maximum or minimum in context.
When do I need the chain rule instead of the power rule?
Use the power rule directly on a single power of \(x\), like \(x^5\). Use the chain rule whenever you're differentiating a function of a function - anything with brackets raised to a power, or a function inside \(\sin\), \(\cos\), \(e^x\) or \(\ln\), such as \((3x+1)^4\) or \(e^{2x}\).
Can I use my GDC to check differentiation on the exam?
Yes, on any calculator paper. Your GDC's numerical derivative feature evaluates \(f'(a)\) at a specific point, which is a fast way to check an algebraic answer or to find a gradient when the algebra would be awkward - but you still need to show the calculus method for full marks. See the GDC guide for model-specific instructions.
Sub-topics
Differentiation broken down into its individual skills, each with its own focused page.
Related topics
More Calculus topics from the same AI HL syllabus unit, in case you want to keep going.